# Problem of hypergeometric testing.

This is actually a computational biology problem, but you don't really need biology knowledge to understand it:

You are given expression values of a given protein P for 10 individuals with a normal condition and 10 individuals with a disease condition.

Normal: 5.2 6.4 7.8 3.1 2.9 1.0 2.3 0.6 4.3 3.2

Disease: 7.8 9.1 10.4 11.5 4.3 6.5 7.6 6.7 10.1 2.1

Your team built a disease predictor such that when the expression of the protein is higher than 5.5, the predictor claims that the individual has the disease. You now want to evaluate out of the disease predictions made by the predictor if the actual number of disease individuals predicted is obtained by chance. Compute the p-value according to the correct statistical test seen in class and assume that a p-value < 0.05 significantly differs from what should be obtained by chance. Show your calculations.

I was given the standard answer that

$$p-value = 1 - \sum_{i=0}^7 \frac {\binom {10} i \binom {20-10}{i}} {\binom {20} {10}}$$

which is using the hypergeometric distribution.

I know what is a $p-value$, but what is hypothesis we are testing with this $p-value$?

For me it seems that the formula is calculating if we pick 10 people from the 20 samples, what's the probability that we get more than 7 normal people. But this probability doesn't seem to have any connection with predictor. I'm confused...

-
You'll have to figure out if "the disease predictor works" is your null or alternative hypothesis... –  Ｊ. Ｍ. Dec 17 '11 at 16:53
What if we assume that "the predictor works" is alternative and "it doesn't works" is null? –  ablmf Dec 17 '11 at 17:02
Yeah, that's a typo. –  ablmf Dec 17 '11 at 23:20

The null hypothesis is that the disease predictor is just guessing at random (in other words, it doesn't work). However, out of the $10$ people who do have the disease, the predictor guessed $10$ and got $8$ of them right. If the predictor is just guessing at random (i.e., assuming the null hypothesis), how likely is it that the predictor would have gotten 8 or more right? The $p$-value is that probability.
Assuming the predictor is just guessing, out of the 10 that the predictor said had the disease, the probability that $i$ of them actually do have the disease is $$\frac{\binom{10}{i}\binom{20-10}{10-i}}{\binom{20}{10}}.$$ (This is the number of ways of choosing $i$ diseased people from the $10$ who are diseased times the number of ways of choosing $10-i$ normal people from the $20-10$ who are normal divided by the total number of ways of choosing $10$ people from the group of $20$. As you noted, this is a probability associated with the hypergeometric distribution.)
Since the $p$-value is the probability that the predictor would have gotten 8 or more right, the $p$-value is $$\sum_{i=8}^{10} \frac{\binom{10}{i}\binom{20-10}{10-i}}{\binom{20}{10}} = 1 - \sum_{i=0}^{7} \frac{\binom{10}{i}\binom{20-10}{10-i}}{\binom{20}{10}} \approx 0.01.$$